If $$\alpha, \beta, \gamma, \delta$$ are the roots of the equation $$x^{4}+x^{3}+x^{2}+x+1=0$$, then $$\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}$$ is equal to :

For $$\mathrm{n} \in \mathbf{N}$$, let $$\mathrm{S}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-3+2 i|=\frac{\mathrm{n}}{4}\right\}$$ and $$\mathrm{T}_{\mathrm{n}}=\left\{z \in \mathbf{C}:|z-2+3 i|=\frac{1}{\mathrm{n}}\right\}$$. Then the number of elements in the set $$\left\{n \in \mathbf{N}: S_{n} \cap T_{n}=\phi\right\}$$ is :

The number of $$\theta \in(0,4 \pi)$$ for which the system of linear equations

$$ \begin{aligned} &3(\sin 3 \theta) x-y+z=2 \\\\ &3(\cos 2 \theta) x+4 y+3 z=3 \\\\ &6 x+7 y+7 z=9 \end{aligned} $$

has no solution, is :

If $$\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$$, then $$8(\alpha+\beta)$$ is equal to :